Question

If the $4^{th},{10}^{th}$ and ${16}^{th}$ terms of a G.P. are $x,y$ and $z$, respectively. Prove that $x,y,z$ are in G.P.

Answer 1

Let $a$ be the first term and $r$ be the common ratio of the G.P.
According to the given condition,
$a_4={ar}^3=x...\left(1\right)a_{10}={ar}^9=y...\left(2\right)a_{16}={ar}^{15}=z...\left(3\right)\text{Dividing (2) by (1), we obtain}\frac{y}{x}=\frac{a{r}^2}{r^3}\Rightarrow \frac{y}{x}=r^6\text{Dividing (3) by (2), we obtain}\frac{z}{y}=\frac{a{r}^{15}}{a{r}^9}\Rightarrow \frac{z}{y}=r^6\therefore \frac{y}{x}=\frac{z}{y}\text{Thus, x, y, z are in G.P.}$